Lecture PowerPoint
Chemistry
The Molecular Nature of
Matter and Change
Seventh Edition
Martin S. Silberberg
and Patricia G. Amateis
23-1 Copyright  McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 23
Transition Elements and Their
Coordination Compounds
23-2
The Transition Elements and Their Coordination
Compounds
23.1 Properties of the Transition Elements
23.2 The Inner Transition Elements
23.3 Coordination Compounds
23.4 Theoretical Basis for the Bonding and Properties of
Complexes
23-3
Figure 23.1
The transition elements (d block) and inner
transition elements (f block) in the periodic table.
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23-4
Properties of the Transition Metals
All transition metals are metals, whereas the main-group
elements in each period comprise both metals and
nonmetals.
Many transition metal compounds are colored and
paramagnetic, whereas most main-group ionic compounds
are colorless and diamagnetic.
The properties of transition metal compounds are
related to the electron configuration of the metal ion.
23-5
Figure 23.2
The Period 4 transition metals.
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23-6
Electron Configurations of
Transition Metals and their Ions
The d-block elements have the general condensed
ground-state configuration [noble gas] ns2(n – 1)dx
where n = 4 to 7 and x = 1 to 10.
Periods 6 and 7 elements include the f sublevel:
[noble gas] ns2(n – 2)f14(n – 1)dx where n = 6 or 7.
Transition metals generally form ions by losing the ns
electrons before the (n – 1)d electrons.
23-7
Table 23.1 Orbital Occupancy of the Period 4 Transition Metals
The number of unpaired electrons increases in the first half of the
series and decreases in the second half, when pairing begins.
23-8
Sample Problem 23.1
Writing Electron Configurations of
Transition Metal Atoms and Ions
PROBLEM: Write condensed electron configurations for the following:
(a) Zr; (b) V3+; (c) Mo3+. (Assume that elements in higher
periods behave like those in Period 4.)
PLAN: We locate the element in the periodic table and count its
position in the respective transition series. These elements are
in Periods 4 and 5, so the general electron configuration is
[noble gas] ns2(n – 1)dx. For the ions, we recall that ns
electrons are generally lost first.
SOLUTION:
(a)
Zr is the second element in the 4d series:
[Kr] 5s24d2
23-9
Sample Problem 23.1
(b) V is the third element in the 3d series, so its configuration is
given by [Ar] 4s23d3. When it forms V3+, it loses the two 4s efirst, then one of the 3d e–:
[Ar] 3d2
(c)
Mo lies below Cr in group 6B(6), so we expect the same
exception as for Cr. The configuration for Mo is therefore given
by [Kr] 5s14d5. Formation of the Mo3+ ion occurs by loss of the
single 5s electron followed by two 4d electrons:
[Kr] 4d3
23-10
Trends in the Properties of
Transition Metals
Across a period the following trends are observed:
Atomic size decreases at first, then remains relatively
constant.
- The d electrons fill inner orbitals, so they shield outer electrons very
efficiently and the 4s electrons are not pulled closer by the increasing
nuclear charge.
Ionization energies also increase relatively little across
the transition metals of a particular period.
Electronegativities tend to increase slightly, then
decrease, across the transition metals in a given period.
23-11
Figure 23.3 Trends in key atomic properties of Period 4 elements.
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23-12
Trends in the Properties of
Transition Metals
Within a group the trends also differ from those observed
for main group elements.
Atomic size increases from Period 4 to 5, but not from
Period 5 to 6.
- A Period 6 element has 32 more protons than its preceding Period
5 group member (compared to a difference of 18 for main-group
elements).
- The extra shrinkage from the increase in nuclear charge (called
the lanthanide contraction) is roughly equal to the normal size
increase due to adding an extra energy level.
23-13
Trends in the Properties of
Transition Metals
Electronegativity increases within a group from Period 4
to 5, then generally remains unchanged from Period 5 to
6. The heavier elements often have high EN values.
Although atomic size increases slightly down the group, nuclear
charge increases much more, leading to higher EN values.
Ionization energy values generally increase down a
transition group, also running counter to the main group
trend.
Density increases dramatically down a group since atomic
volumes change little while atomic masses increase
significiantly.
23-14
Figure 23.4
23-15
Vertical trends in key properties within the
transition elements.
Oxidation States of Transition Metals
Most transition metals have multiple oxidation states.
The highest oxidation state for elements in Groups 3B(3)
through 7B(7) equals the group number.
- These states are seen when the elements combine with the highly
electronegative oxygen or fluorine.
Elements in Groups 8B(8), 8B(9), and 8B(10) exhibit
fewer oxidation states. The higher oxidation state is less
common and never equal to the group number.
- The +2 oxidation state is common because the ns2 electrons
are readily lost.
23-16
Figure 23.5
23-17
Aqueous oxoanions of transition elements.
Table 23.2 Oxidation States and d-Orbital Occupancy of the
Period 4 Transition Metals*
23-18
Metallic Behavior of Transition Metals
The lower the oxidation state of the transition metal, the
more metallic its behavior.
Ionic bonding is more prevalent for the lower oxidation
states, whereas covalent bonding occurs more
frequently for higher oxidation states.
Metal oxides become less basic (more acidic) as the
oxidation state increases.
A metal atom in a positive oxidation state has a greater attraction
for bonded electrons, and therefore a greater effective
electronegativity, or valence-state electronegativity, than in the
zero oxidation state. This effect increases as its oxidation state
increases.
23-19
Table 23.3 Standard Electrode Potentials of Period 4 M2+ Ions
Half-Reaction
E°(V)
Ti2+(aq) + 2e–
Ti(s)
−1.63
V2+(aq) + 2e–
V(s)
−1.19
Cr2+(aq) + 2e–
Cr(s)
−0.91
Mn2+(aq) + 2e–
Mn(s)
−1.18
Fe2+(aq) + 2e–
Fe(s)
−0.44
Co2+(aq) + 2e–
Co(s)
−0.28
Ni2+(aq) + 2e–
Ni(s)
−0.25
Cu2+(aq) + 2e–
Cu(s)
0.34
Zn2+(aq) + 2e–
Zn(s)
−0.76
23-20
In general, reducing strength
decreases across the series.
Color and Magnetic Behavior
Most main-group ionic compounds are colorless and
diamagnetic because the metal ion has no unpaired
electrons.
Many transition metal ionic compounds are highly
colored and paramagnetic because the metal ion has
one or more unpaired electrons.
Transition metal ions with a d0 or d10 configuration are
also colorless and diamagnetic.
23-21
Figure 23.6
23-22
Colors of representative compounds of the Period 4
transition metals.
Table 23.4 Some Properties of Group 6B(6) Elements
Atomic Radius
Element
(pm)
IE1(kJ/mol)
E° (V) for
M3+(aq)/M(s)
Cr
128
653
−0.74
Mo
139
685
−0.20
W
139
770
−0.11
IE1 increases down the group, so reactivity decreases. This
trend is opposite to that seen in main-group elements.
23-23
Lanthanides and Actinides
The lanthanides are also called the rare earth elements.
The atomic properties of the lanthanides vary little
across the period, and their chemical properties are
also very similar.
Most lanthanides have the ground-state electron
configuation [Xe] 6s24fx5d0.
All actinides are radioactive, and have very similar physical
and chemical properties.
The +3 oxidation state is common for both lanthanides
and actinides.
23-24
Sample Problem 23.2
Finding the Number of Unpaired Electrons
PROBLEM: The alloy SmCo5 forms a permanent magnet because
both samarium and cobalt have unpaired electrons. How
many unpaired electrons are in Sm (Z = 62)?
PLAN: We write the condensed electron configuration of Sm and
then, using Hund’s rule and the aufbau principle, place
electrons into a partial orbital diagram and count the
unpaired electrons.
SOLUTION:
Sm is the eighth element after Xe. Two electrons go into the 6s
sublevel. In general, the 4f sublevel fills before the 5d, so the
remaining six electrons go into the 4f sublevel.
The condensed configuration of Sm is [Xe] 6s24f6.
23-25
Sample Problem 23.2
The partial orbital diagram is:
↑↓
6s
↑
↑
↑
↑
4f
↑
↑
5d
Sm has six unpaired electrons.
23-26
6p
Coordination Compounds
A coordination compound contains at least one complex
ion, which consists of a central metal cation bonded to
molecules and/or anions called ligands.
The complex ion is associated with counter ions of
opposite charge.
The complex ion [Cr(NH3)6]3+ has a central Cr3+ ion bonded to six
NH3 ligands. The complex ion behaves like a polyatomic ion in
solution.
23-27
Coordination Number
The coordination number is the number of ligand atoms
bonded directly to the central metal ion.
Coordination number is specific for a given metal ion in a
particular oxidation state and compound.
- [Cr(NH3)6]3+ has a coordination number of 6.
The most common coordination number in complex ions is
6, but 2 and 4 are often seen.
23-28
Figure 23.7
Components of a coordination compound.
[Co(NH3)6]Cl3 dissolves in water. The six
ligands remain bound to the complex ion.
23-29
K2[Pt(Cl)4] has four Cl–
ligands and 2 K+ counter
ions.
Table 23.5 Coordination Numbers and Shapes of Some
Complex Ions
Coordination
Number
Shape
Examples
2
Linear
[CuCl2] –, [Ag(NH3)2]+, [AuCl2] –
4
Square planar
[Ni(CN)4]2–, [PdCl4]2–,
[Pt(NH3)4]2+, [Cu(NH3)4]2+
4
Tetrahedral
[Cu(CN)4]3–, [Zn(NH3)4]2+,
[CdCl4]2–, [MnCl4]2–
6
Octahedral
[Ti(H2O)6]3+, [V(CN)6]4–,
[Cr(NH3)4Cl2]+, [Mn((H2O6]2+,
[FeCl6]3–, [Co(en)3]3+
The geometry of a given complex ion depends both on the
coordination number and the metal ion.
23-30
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Ligands
The ligands of a complex ion are molecules or anions
with one or more donor atoms.
Each donor atom donates a lone pair of electrons to
the metal ion to form a covalent bond.
Ligands are classified in terms of their number of donor
atoms, or “teeth”:
- Monodentate ligands bond through a single donor atom.
- Bidentate ligands have two donor atoms, each of which bonds to
the metal ion.
- Polydentate ligands have more than two donor atoms.
23-31
Table 23.6 Some Common Ligands in Coordination Compounds
23-32
Chelates
Bidentate and polydentate ligands give rise to rings in
the complex ion.
A complex ion containing this type of structure is called a
chelate because the ligand seems to grab the metal ion
like claws.
EDTA has six donor atoms and forms very stable complexes
with metal ions.
23-33
Formulas of Coordination Compounds
• A coordination compound may consist of
– a complex cation with simple anionic counterions,
– a complex anion with simple cationic counterions, or
– a complex cation with complex anion as counterion.
• When writing the formula for a coordination compound
– the cation is written before the anion,
– the charge of the cation(s) is/are balanced by the charge of the
anion(s), and
– neutral ligands are written before anionic ligands, and the
formula of the whole complex ion is placed in square brackets.
23-34
Determining the Charge of the Metal Ion
The charge of the cation(s) is/are balanced by the charge
of the anion(s).
K2[Co(NH3)2Cl4] contains a complex anion.
The charge of the anion is balanced by the two K+ counter ions, so
the anion must be [Co(NH3)2Cl4]2–.
There are two neutral NH3 ligands and four Cl- ligands. To have an
overall charge of 2-, the metal ion must have a charge of 2+.
Charge of complex ion = charge of metal ion + total charge of ligands
2= charge of metal ion + [(2 x 0) + (4 x -1)]
Charge of metal ion = (-2) – (-4) = +2 or 2+
The metal ion in this complex anion is Co2+.
23-35
[Co(NH3)4Cl2]Cl contains a complex cation.
The charge of the cation is balanced by the Cl- counter ion, so the
cation must be [Co(NH3)4Cl2]+.
There are four neutral NH3 ligands and two Cl- ligands. To have an
overall charge of 1+, the metal ion must have a charge of 3+.
Charge of complex ion = charge of metal ion + total charge of ligands
1+
= charge of metal ion + [(4 x 0) + (2 x -1)]
Charge of metal ion = (+1) – (-2) = +3 or 3+
The metal ion in this complex cation is Co3+.
23-36
Sample Problem 23.3
Finding the Coordination Number and Metal
Ion Charge in Coordination Compounds
PROBLEM: Give the coordination number and charge of the central metal
ion in the following coordination compounds: (a) Na2[Zn(OH)4];
(b) K[Co(H2O)2(C2O4)2]; (c) [Ru(H2O)2(NH3)2Cl2]Br.
PLAN: To find the coordination number, we use Table 23.6 to determine the
number of ligand atoms bonded to the central metal ion in the complex ion
(in square brackets). Counter ions (those outside the square brackets) are
not included in the determination of the coordination number. We know that
the charge of the complex ion must balance the charge of the counter ions;
the charge of the metal ion is equal to the charge of the complex ion minus
the total charge of the ligands (from Table 23.2).
SOLUTION: (a) Each of the four OH− monodentate ligands forms one bond
to the metal ion for a coordination number of 4. Since the two Na+ counter
ions have a total charge of 2+, the charge of the complex ion is 2-; the four
OH− ligands have a total charge of 4-.
The charge of the central metal (zinc) ion
= charge of complex ion – total charge of ligands
= (2-) – (4-) = 2+
23-37
Sample Problem 23.3
SOLUTION: (b) There are two monodentate H2O ligands, each forming one
bond to the metal ion; since the two C2O42− ligands are bidentate, each of
these ligands forms two bonds to the metal ion, for a total of four bonds.
The coordination number is 6. Since the K+ counter ion has a charge of 1+,
the charge of the complex ion is 1-. The H2O ligands are neutral and the
two C2O42− ligands have a total charge of 4-.
The charge of the central metal (cobalt) ion = (1-) – [0 + (4-)] = 3+
(c) The complex has two H2O, two NH3, and two Cl− ligands, all of which
form one bond to the metal ion for a coordination number of 6. Since the
Br– counter ion has a 1- charge, the complex ion has a charge of 1+. The
H2O and NH3 ligands are neutral and the two Cl− ligands have a total
charge of 2-.
The charge of the metal (ruthenium) ion = (1+) – [0 + 0 + (2-)] = 3+
CHECK: Be sure that the coordination number equals the sum of the
number of monodentate ligands and twice the number of bidentate ligands.
Be sure that the total charge of the metal ion, ligands, and counter ions
equals zero.
23-38
Naming Coordination Compounds
• The cation is named before the anion.
• Within the complex ion, the ligands are named in alphabetical order
before the metal ion.
– Anionic ligands drop the –ide and add –o after the root name.
• A numerical prefix is used to indicate the number of ligands of a
particular type.
– Prefixes do not affect the alphabetical order of ligand names.
– Ligands that include a numerical prefix in the name use the
prefixes bis (2), tris (3), or tetrakis (4) to indicate their number.
• A Roman numeral is used to indicate the oxidation state for a metal
that can have more than one state.
• If the complex ion is an anion, we drop the ending of the metal name
and add –ate.
23-39
Table 23.7 Names of Some Neutral and Anionic Ligands
Neutral
23-40
Anionic
Name
Formula
Name
Formula
Aqua
H 2O
Fluoro
F–
Ammine
NH3
Chloro
Cl–
Carbonyl
CO
Bromo
Br–
Nitrosyl
NO
Iodo
I–
Hydroxo
OH–
Cyano
CN–
Table 23.8 Names of Some Metal Ions in Complex Anions
23-41
Metal
Name in Anion
Iron
Ferrate
Copper
Cuprate
Lead
Plumbate
Silver
Argentate
Gold
Aurate
Tin
Stannate
Sample Problem 23.4
Writing Names and Formulas of
Coordination Compounds
PROBLEM:
(a) What is the systematic name of Na3[AlF6]?
(b) What is the sytematic name of [Co(en)2Cl2]NO3?
(c) What is the formula of tetraamminebromochloroplatinum(IV) chloride?
(d) What is the formula of hexaamminecobalt(III) tetrachloroferrate(III)?
PLAN: We use the rules for writing formulas and names of
coordination compounds.
SOLUTION:
(a) The complex ion is [AlF6]3–. There are six (hexa-) F– ions (fluoro)
as ligands. The complex ion is an anion, so the ending of the
metal name must be changed to –ate. Since Al has only one
oxidation state, no Roman numerals are used.
sodium hexafluoroaluminate
23-42
Sample Problem 23.4
(b) There are two ligands, Cl– (chloro) and en (ethylenediamine).
The ethylenediamine ligand already has a numerical prefix in
its name, so we indicate the two en ligands by the prefix bis
instead of di.
The complex ion is a cation, so the metal name is unchanged,
but we need to specify the oxidation state of Co. The counter
ion is NO3–, so the complex ion is [Co(en)2Cl2]+.
Charge of complex ion = charge of metal ion + total charge of ligands
1+
= charge of metal ion + [(2 x 0) + (2 x 1-)]
Charge of metal ion = (+1) – (-2) = +3 or 3+
The ligands must be named in alphabetical order:
dichlorobis(ethylenediamine)cobalt(III) nitrate
23-43
Sample Problem 23.4
(c) The name of the central metal ion is written first, followed by
the neutral ligands and then (in alphabetical order) by the
negative ligands.
Charge of complex ion = charge of metal ion + total charge of ligands
= (4+) + [(4 x 0) + (1 x 1-) + (1 X 1-)]
= +4 + (-2) = +2 or 2+
We will therefore need two Cl- counter ions to balance the charge on the
complex ion.
[Pt(NH3)4BrCl]Cl2
23-44
Sample Problem 23.4
(d) This compound consists of two different complex ions. In the
cation, there are six NH3 ligands and the metal ion is Co3+, so
the cation is [Co(NH3)6]3+.
The anion has four Cl– ligands and the central metal ion is Fe3+,
so the ion is [FeCl4] –.
The charge on the cation must be balanced by the charge on
the anion, so we need three anions for every one cation:
[Co(NH3)6][FeCl4]3
23-45
Constitutional Isomers of Coordination
Compounds
Compounds with the same formula, but with the atoms
connected differently, are constitutional isomers.
Coordination isomers occur when the composition of
the complex ion, but not the compound, is different.
- This can occur by the exchange of a ligand and a counter ion, or by
the exchange of ligands.
Linkage isomers occur when the composition of the
complex ion is the same but the ligand donor atom is
different.
- Some ligands can bind to the metal through either of two donor
atoms.
23-46
Figure 23.8 A pair of linkage (constitutional) isomers
The nitrite ion can bind either through the N atom or either one of
the O atoms.
23-47
Ligands that have more than one donor atom
23-48
Stereoisomers of Coordination
Compounds
Stereoisomers are compounds that have the same atomic
connections but different spatial arrangements of their
atoms.
Geometric or cis-trans isomers occur when atoms or
groups can either be arranged on the same side or on
opposite sides of the compound relative to the central
metal ion.
Optical isomers (enantiomers) are nonsuperimposable
mirror images of each other.
23-49
Figure 23.9A
Geometric (cis-trans) isomerism.
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The cis and trans isomers of [Pt(NH3)2Cl2].
In the cis isomer, identical ligands are adjacent to each other, while in
the trans isomer they are across from each other.
The cis isomer (cisplatin) is an antitumor agent while the trans
isomer has no antitumor effect.
23-50
Figure 23.9B
Geometric (cis-trans) isomerism.
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The cis and trans isomers of [Co(NH3)4Cl2]+.
Note the placement of the Cl– ligands (green spheres).
23-51
Figure 23.10AB
23-52
Optical isomerism in an octahedral complex ion.
Figure 23.11
23-53
Important types of isomerism in coordination
compounds.
Sample Problem 23.5 Determining the Type of Stereoisomerism
PROBLEM: Draw stereoisomers for each of the following and state the
type of isomerism each exhibits:
..
..
3+
(a) [Pt(NH3)2Br2] (square planar) (b) [Cr(en)3] (en = H2NCH2CH2NH2)
PLAN: We determine the geometry around each metal ion and the
nature of the ligands. If there are different ligands that can be
placed in different positions relative to each other, geometric
(cis-trans) isomerism occurs. Then we see whether the mirror
image of an isomer is superimposable on the original. If it is
not, optical isomerism also occurs.
23-54
Sample Problem 23.5
SOLUTION:
(a) The square planar Pt(II) complex has two different types of
monodentate ligands. Each pair of ligands can be next to
each other or across from each other. Thus geometric
isomerism occurs.
Br
NH3
H3 N
Pt
H3N
Pt
Br
trans
NH3
Br
Br
cis
These are geometric isomers; they do not have optical isomers
since each compound is superimposable on its mirror image.
23-55
Sample Problem 23.5
(b) Ethylenediamine (en) is a bidentate ligand. The Cr3+ ion has
a coordination number of 6 and an octahedral geometry, like
Co3+. The three bidentate ligands are identical, so there is
no geometric isomerism. However, the complex ion has a
nonsuperimposable mirror image. Thus optical isomerism
occurs.
3+
3+
N
N
Cr
N
N
N
N
N
N
N
Cr
N
N
N
not the same
mirror
rotate 180º in
square plane about
vertical axis
3+
N
N
Cr
N
N
N
23-56
N
Bonding in Complex Ions
When a complex ion is formed, each ligand donates an
electron pair to the metal ion.
The ligand acts as a Lewis base, while the metal ion acts as a Lewis
acid.
This type of bond is called a coordinate covalent bond
since both shared e- originate from one atom in the pair.
In terms of valence bond theory, the filled orbital of the
ligand overlaps with an empty orbital of the metal ion.
The VB model proposes that the geometry of the complex
ion depends on the hybridization of the metal ion.
23-57
Figure 23.12
23-58
Hybrid orbitals and bonding in the octahedral
[Cr(NH3)6]3+ ion.
Figure 23.13
23-59
Hybrid orbitals and bonding in the square planar
[Ni(CN)4]2- ion.
Figure 23.14 Hybrid orbitals and bonding in the tetrahedral
[Zn(OH)4]2- ion.
23-60
Figure 23.15
An artist’s wheel.
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Each color has a complementary
color; the one opposite it on the
artist’s wheel.
The color an object exhibits depends
on the wavelengths of light that it
absorbs.
An object will have a particular color because
• it reflects light of that color, or
• it absorbs light of the complementary color.
23-61
Table 23.9 Relation Between Absorbed and Observed Colors
Absorbed Color
l (nm)
Observed Color
l (nm)
Violet
400
Green-yellow
560
Blue
450
Yellow
600
Blue-green
490
Red
620
Yellow-green
570
Violet
410
Yellow
580
Dark blue
430
Orange
600
Blue
450
Red
650
Green
520
23-62
Crystal Field Theory
Crystal field theory explains color and magnetism in
terms of the effect of the ligands on the energies of the
d-orbitals of the metal ion.
The bonding of the ligands to the metal ion cause the
energies of the metal ion d-orbitals to split.
Although the d-orbitals of the unbonded metal ion are equal in
energy, they have different shapes, and therefore different
interactions with the ligands.
The splitting of the d-orbitals depends on the relative
orientation of the ligands.
23-63
Figure 23.16
23-64
The five d-orbitals in an octahedral field of ligands.
Figure 23.17
Splitting of d-orbital energies in an octahedral
field of ligands.
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The d orbitals split into two groups. The difference in energy
between these groups is called the crystal field splitting energy,
symbol D.
23-65
Figure 23.18
The effect of ligands and splitting energy on
orbital occupancy.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Weak field ligands lead to
a smaller splitting energy.
23-66
Strong field ligands lead to a
larger splitting energy.
Figure 23.19
The color of [Ti(H2O)6]3+.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
The hydrated Ti3+ ion is purple.
Green and yellow light are absorbed
while other wavelengths are
transmitted. This gives a purple color.
23-67
Figure 23.19
The color of [Ti(H2O)6]3+.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
When the ion absorbs light, electrons can move from the lower t2g
energy level to the higher eg level. The difference in energy
between the levels (D) determines the wavelengths of light
absorbed. The visible color is given by the combination of the
wavelengths transmitted.
23-68
The Colors of Transition Metal Complexes
The color of a coordination compound is determined
by the crystal field splitting energy (D) of its complex
ion.
For a given ligand, the color depends on the oxidation
state of the metal ion.
For a given metal ion, the color depends on the ligand.
23-69
Figure 23.20
Effects of oxidation state and ligand on color.
A change in oxidation state
causes a change in color.
Substitution of an NH3 ligand
with a Cl– ligand affects the
color of the complex ion.
23-70
Figure 23.21
The spectrochemical series
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I- < Cl- < F- < OH- < H2O < SCN- < NH3 < en < NO2- < CN- < CO
WEAKER FIELD
STRONGER FIELD
SMALLER D
LARGER D
LONGER l
SHORTER l
As D increases, shorter wavelengths (higher energies) of light
must be absorbed to excite electrons. For reference H2O is
considered a weak-field ligand.
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Sample Problem 23.6
Ranking Crystal Field Splitting Energies (Δ)
for Complex Ions of a Metal
PROBLEM: Rank the ions [Ti(H2O)6]3+, [Ti(NH3)6]3+, and [Ti(CN)6]3– in
terms of D and of the energy of visible light absorbed.
PLAN: The formulas show that Ti has an oxidation state of +3 in all
three ions. From Figure 23.21, we rank the ligands by crystal
field strength: the stronger the ligand, the greater the
splitting, and the greater the energy of light absorbed.
SOLUTION: The ligand field strength is CN– > NH3 > H2O, so the
relative size of D and energy of light absorbed will be
[Ti(CN)6]3– > [Ti(NH3)6]3+ > [Ti(H2O)6]3+
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The Magnetic Properties of Transition
Metal Complexes
Magnetic properties are determined by the number of
unpaired electrons in the d orbitals of the metal ion.
Hund’s rule states that e- occupy orbitals of equal energy
one at a time. When all lower energy orbitals are halffilled:
- The next e- can enter a half-filled orbital and pair up by
overcoming a repulsive pairing energy, (Epairing), or
- the next e- can enter an empty, higher, energy orbital by
overcoming D.
The number of unpaired e- will depend on the relative
sizes of Epairing and D.
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Figure 23.22
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High-spin and low-spin octahedral complex
ions of Mn2+.
Figure 23.23 Orbital occupancy for high-spin and low-spin
octahedral complexes of d4 through d7 metal ions.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
high spin:
weak-field
ligand
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low spin:
strong-field
ligand
high spin:
weak-field
ligand
low spin:
strong-field
ligand
Sample Problem 23.7 Identifying High-Spin or Low-Spin Complex
Ions
PROBLEM: Iron(II) forms a complex in hemoglobin. For each of
the two octahedral complex ions [Fe(H2O)6]2+ and
[Fe(CN)6]4–, draw an energy diagram showing orbital
splitting, predict the number of unpaired electrons, and
identify the ion as low spin or high spin.
PLAN:
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The Fe2+ electron configuration shows the number of d
electrons, and the spectrochemical series shows the
relative ligand strengths. We draw energy diagrams and
separate the t2g and eg orbital sets more for the strong-field
ligand. Then we add electrons, noting that a weak-field
ligand gives the maximum number of unpaired electrons
and a high-spin complex, whereas the strong-field ligand
will give the minimum number of unpaired electrons and a
low-spin complex.
Sample Problem 23.6
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↑
↑↓ ↑
eg
↑
eg
↑
t2g
Potential energy
Potential energy
SOLUTION:
↑↓ ↑↓ ↑↓ t2g
[Fe(H2O)6]2+
[Fe(CN)6]4-
high-spin
low-spin
Figure 23.24
Splitting of d-orbital energies by a (A) tetrahedral
and (B) square planar field of ligands.
The splitting of d-orbital energies is less in a
tetrahedral than an octahedral complex, and
the relative d-orbital energies are
reversed. Only high-spin tetrahedral
complexes are known because D is small.
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Square planar complexes are low-spin
and usually diamagnetic because the
four pairs of d electrons fill the four
lowest-energy orbitals.
Chemical Connections
Figure B23.1
Hemoglobin and the octahedral complex in heme.
Hemoglobin consists of four protein chains, each with a bound
heme. In oxyhemoglobin (B), the octahedral complex in heme has
an O2 molecule as the sixth ligand for iron(II).
(Illustration by Irving Geis. Rights owned by Howard Hughes Medical Institute. Not to be used without permission.)
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Table B23.1 Some Transition Metal Trace Elements in Humans
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Chemical Connections
Figure B23.2
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The tetrahedral Zn2+ complex in carbonic anhydrase.